# A projectile is shot from the ground at an angle of pi/12  and a speed of 7 /18 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Jul 19, 2017

distance from starting point is $\frac{5}{1296} m \approx 3.858 \text{x} {10}^{-} 3 m$

#### Explanation:

Range, $R = {V}^{2} / g \sin \left(2 \theta\right)$

When the particle reaches its maximum height, the distance from starting point will be $\frac{R}{2}$

$\frac{R}{2} = {V}^{2} / \left(2 g\right) \sin \left(2 \theta\right)$ where V is the initial launching speed

$\frac{R}{2} = {\left(\frac{7}{18}\right)}^{2} / \left(2 g\right) \sin \left(2 \cdot \frac{\pi}{12}\right)$

$\frac{R}{2} = \frac{\frac{49}{324}}{2 g} \sin \left(\frac{\pi}{6}\right)$

[Take $g = 9.8 m {s}^{-} 2$]

$\frac{R}{2} = \left(\frac{5}{648}\right) \sin \left(\frac{\pi}{6}\right)$

$\frac{R}{2} = \frac{5}{1296} m \approx 3.858 \text{x} {10}^{-} 3 m$

Jul 19, 2017

$\text{distance} = 0.00389$ $\text{m}$

#### Explanation:

We're asked to find the distance from the launch point a projectile is when it reaches its maximum height.

We can use the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0 y}\right)}^{2} - 2 g h$

to find the maximum height $h$.

When the particle is at its maximum height, the instantaneous velocity ${v}_{y}$ is $0$.

The initial $y$-velocity ${v}_{0 y}$ is

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0} = \left(\frac{7}{18} \textcolor{w h i t e}{l} \text{m/s}\right) \sin \left(\frac{\pi}{12}\right) = 0.101$ $\text{m/s}$

Plugging in known values, we have

$0 = \left(0.101 \textcolor{w h i t e}{l} {\text{m/s")^2 - 2(9.81color(white)(l)"m/s}}^{2}\right) h$

h = ((0.101color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(5.16xx10^-4 color(red)("m"

To find the horizontal distance covered $\Delta x$, we use the equation

$\Delta x = {v}_{0 x} t$

${v}_{0 x} = {v}_{0} \cos {\alpha}_{0} = 0.376$ $\text{m/s}$

The time $t$ when it reaches its maximum height is given by

${v}_{y} = {v}_{0 y} - g t$

Rearranging this equation and plugging in known values:

$t = \frac{{v}_{0 y} - {v}_{y}}{g} = \left(0.101 \textcolor{w h i t e}{l} {\text{m/s" - 0)/(9.81color(white)(l)"m/s}}^{2}\right) = 0.0103$ $\text{s}$

Thus, we have

Deltax = (0.376color(white)(l)"m/s")(0.0103color(white)(l)"s") = color(green)(0.00385 color(green)("m"

The distance from the starting point is thus

$r = \sqrt{{\left(\Delta x\right)}^{2} + {h}^{2}} = \sqrt{{\left(\textcolor{g r e e n}{0.00385 \textcolor{w h i t e}{l} \text{m"))^2 + (color(red)(5.16xx10^-4color(white)(l)"m}}\right)}^{2}}$

= color(blue)(0.00389 color(blue)("m"

Jul 19, 2017

THe distance is $= 0.0039 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

initial velocity is ${u}_{y} = v \sin \theta = \frac{7}{18} \cdot \sin \left(\frac{1}{12} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = \frac{7}{18} \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{7}{18 g} \cdot \sin \left(\frac{1}{12} \pi\right)$

$= 0.0103 s$

The greatest height is

$h = {\left(\frac{7}{18} \sin \left(\frac{1}{12} \pi\right)\right)}^{2} / \left(2 g\right) = 0.00052 m$

Resolving in the horizontal direction ${\rightarrow}^{+}$

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= \frac{7}{18} \cos \left(\frac{1}{12} \pi\right) \cdot 0.0103$

$= 0.00387 m$

The distance from the starting point is

$d = \sqrt{{h}^{2} + {s}^{2}}$

$= \sqrt{{0.00052}^{2} + {0.00387}^{2}}$

$= 0.0039 m$